If α,β,γ,δ∈R satisfy α+β+γ+δ(α+1)2+(β+1)2+(γ+1)2+(δ+1)2=4 If biquadratic equation a0x4+a1x3+a2x2+a3x+a4=0 has the roots (α+β1−1),(β+γ1−1),(γ+δ1−1),(δ+α1−1). Then the value of a0a2 is :(a) 4(b) −4(c) 6(d) none of these
Q. If α,β,γ,δ∈R satisfy α+β+γ+δ(α+1)2+(β+1)2+(γ+1)2+(δ+1)2=4 If biquadratic equation a0x4+a1x3+a2x2+a3x+a4=0 has the roots (α+β1−1),(β+γ1−1),(γ+δ1−1),(δ+α1−1). Then the value of a0a2 is :(a) 4(b) −4(c) 6(d) none of these
Expand and Simplify: Given the equation:α+β+γ+δ(α+1)2+(β+1)2+(γ+1)2+(δ+1)2=4We can simplify this equation by expanding the squares and then dividing by the sum of α, β, γ, and δ.
Divide by Sum: Expanding the numerator, we get: (α2+2α+1)+(β2+2β+1)+(γ2+2γ+1)+(δ2+2δ+1)This simplifies to:(α2+β2+γ2+δ2)+2(α+β+γ+δ)+4
Set up Equation: Now, we divide by the sum of α, β, γ, and δ: α+β+γ+δ(α2+β2+γ2+δ2)+2(α+β+γ+δ)+4 This simplifies to: (α+β+γ+δ)+2+α+β+γ+δ4 Given that the whole expression equals 4, we can set up the equation: (α+β+γ+δ)+2+α+β+γ+δ4=4
Solve for S: We can now solve for (α+β+γ+δ):(α+β+γ+δ)+(α+β+γ+δ)4=2Let's denote S=α+β+γ+δ. Then we have:S+S4=2Multiplying both sides by S to clear the fraction, we get:S2+4=2S
Quadratic Equation: Rearranging the terms, we get a quadratic equation in S: S2−2S+4=0This equation does not have real roots, which means there is a math error in our previous steps. We need to re-evaluate our approach to solving for S.
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